Integrated Rate Expressions and Working Equations

Equation (8) can be rearranged and integrated as follows:

ln C = -kt + constant where ln C is the natural logarithm (base e) of the concentration.

The constant in equation (10) can be evaluated at zero time when kt = 0 and C = C0, the initial concentration. Thus, ln C0 = constant and log C = -kt + ln C0 (11)

Equation (11) is the integrated rate expression for a first-order process and can serve as a working equation for solving problems. It is also in the form of the equation of a straight line:

Therefore, if ln C is plotted against t, as shown in Figure 2, the plot will be a straight line with an intercept (at t = 0) of ln C0, and the slope of the line (m) will be —k. Such plots are commonly used to determine the order of a reaction; that is, if a plot of ln C versus time is a straight line, the reaction is assumed to be a first-order or pseudo-first-order process.

The slope of the line and the corresponding value of k for a plot such as that shown in Figure 2 may be calculated using the following equation:

Example. A solution of an ester in pH 10.0 buffer (25°C) one hour after preparation was found to contain 3 mg/mL. Two hours after preparation, the solution contained 2 mg/mL. Calculate the psuedo-first-order rate constant for hydrolysis of the ester at pH 10.0 (25°C).

slope(m)

Note that since ln C is dimensionless, the rate constant, k, has the dimensions of reciprocal time (i.e., day-1, hr-1, min-1, sec-1).

Another useful working equation can be obtained by rearranging equation (11) as follows:

Equation (13) shows that since k is a constant for a given process, the value of t is determined solely by the ratio C0/C. For example, when C0/C is equal to 2 the value t will be the same, no matter what the value of the initial concentration (C0) was.

Example. For the ester hydrolysis above (k would C = 1.5 mg/mL?

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